Question 186871
Let {{{r}}}= the speed of the vehicle
Let {{{d}}}= distance in miles 
Let {{{t}}} = time in hours
given:
{{{d = 150}}} mi
{{{d = r*t}}}
(1) {{{150 = r*t}}}
(2) {{{150 = (r + 20)*(t - 2)}}}
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From (1),
{{{t = 150/r}}}
Substitute this in (2)
(2) {{{150 = (r + 20)*((150/r) - 2)}}}
(2) {{{150 = 150 + 3000/r - 2r - 40}}}
Subtract {{{150}}} from both sides
(2) {{{0 = 3000/r - 2r - 40}}}
Add {{{40}}} to both sides
(2) {{{40 = 3000/r - 2r}}}
Multiply both sides by {{{r}}}
(2) {{{40r = 3000 - 2r^2}}}
Add {{{2r^2}}} to both sides
(2) {{{2r^2 + 40r = 3000}}}
divide both sides by {{{2}}}
(2) {{{r^2 + 20r = 1500}}}
I will find {{{r}}} by completing the square 
(you could use quadratic formula also)
(2) {{{r^2 + 20r  + (20/2)^2 = 1500 + (20/2)^2}}}
(2) {{{r^2 + 20r  + 100 = 1500 + 100}}}
(2) {{{r^2 + 20r  + 100 = 1600}}}
Now both sides are perfect squares
(2) {{{(r + 10)^2 = 40^2}}}
Take the square roots of both sides
(2) {{{r + 10 = 40}}}
(2) {{{r = 30}}} (there is a negative solution also, {{{r = -50}}})
The speed of his vehicle is 30 mi/hr
check answer:
(1) {{{150 = r*t}}}
(1) {{{t = 150/30}}}
(1) {{{t = 5}}} hrs
and
(2) {{{150 = (r + 20)*(t - 2)}}}
(2) {{{150 = (30 + 20)*(5 - 2)}}}
(2) {{{150 = 50*3}}}
(2) {{{150 = 150}}}
OK